Cmr S=3+3^2+3^3+3^4+…+3^2012 chia hết cho 40

Question

Cmr S=3+3^2+3^3+3^4+…+3^2012 chia hết cho 40

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Jasmine 2 tháng 2021-10-05T09:28:17+00:00 2 Answers 9 views 0

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    0
    2021-10-05T09:30:14+00:00

    Đáp án : mẫu nha cách lm giống như vậy

    A=3+3^2+3^3+…+3^100

    A=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+…+(3^97+3^98+3^99+3^100)

    A=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+….+3^97(1+3+3^2+3^3)

    A=3.40+3^5.40+….+3^97.40

    A=40.(3+3^5+…+3^97)chia hết cho 40

    Vậy A chia hết cho 40

    0
    2021-10-05T09:30:14+00:00

    Đáp án:

    Giải thích các bước giải:

    $S=3+3^2+3^3+….+3^{2012}$

    $=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+…+(3^{2009}+3^{2010}+3^{2011}+3^{2012})$

    $=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+….+3^{2009}(1+3+3^2+3^3)$

    $=40(3+3^5+…+3^{2009})$

    Do đó S chia hết cho 40

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